Class prep · TJ AP Calculus AB

Learn AB calculus from zero,and know why each step works.

A full AP Calculus AB course for current TJ students, usually taken in 11th or 12th grade, taught with worked examples and original practice. It covers the AP Calculus AB topics: limits and continuity, the derivative and its rules, composite, implicit and inverse functions, applications of differentiation, integration and accumulation of change, differential equations, and applications of integration. Students taking AP Calculus BC should use the AP Calculus BC course instead. Every problem has a full worked solution. Built by Neil Kothari, a TJ alumnus.

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$395 covers the full school year. The first lesson is free.

Try one. This is a test-level problem from the course.

Problem 1

Test level
Multiple choice
No calculator
4 pts

Functions $f$ and $g$ satisfy $\displaystyle\lim_{x\to1^-}f(x)=3$ , $\displaystyle\lim_{x\to1^+}f(x)=-1$ , $f(1)=2$ , $\displaystyle\lim_{x\to1^-}g(x)=-1$ , $\displaystyle\lim_{x\to1^+}g(x)=3$ , and $g(1)$ is undefined. Which of the following limits exist? $\text{I. }\lim_{x\to1}\big[f(x)+g(x)\big]\qquad \text{II. }\lim_{x\to1}\big[f(x)\,g(x)\big]\qquad \text{III. }\lim_{x\to1}\frac{f(x)}{g(x)}$

Answer

I and II only. Sum: $3+(-1)=2$ from the left and $-1+3=2$ from the right. Product: $3(-1)=-3$ and $(-1)(3)=-3$ . Quotient: $\frac{3}{-1}=-3$ and $\frac{-1}{3}=-\frac13$ , which differ.

Full solution

Neither $f$ nor $g$ has a two-sided limit at $1$ , so the limit laws (which need each limit to exist) do not apply. Fall back on one-sided limits: each one-sided limit does exist, and one-sided limits obey the sum, product and quotient laws.
The rule that saves you: combine left with left and right with right. Never mix a left-hand value of $f$ with a right-hand value of $g$ .
I. From the left: $3+(-1)=2$ . From the right: $-1+3=2$ . They agree, so $\lim_{x\to1}[f(x)+g(x)]=2$ exists. Graphically, $f$ jumps down by $4$ exactly where $g$ jumps up by $4$ , so the jumps cancel.
II. From the left: $3\cdot(-1)=-3$ . From the right: $(-1)\cdot3=-3$ . They agree, so $\lim_{x\to1}f(x)g(x)=-3$ exists.
III. From the left: $\frac{3}{-1}=-3$ . From the right: $\frac{-1}{3}=-\frac13$ . Neither denominator limit is $0$ , so each one-sided limit is fine, but they differ. The quotient's limit does not exist.
So exactly I and II exist: choice (B). The values $f(1)=2$ and $g(1)$ undefined are distractions; they play no role in any of these limits.
Why the other choices tempt people. (E) comes from believing that if neither part has a limit, no combination can. (A) comes from trusting the sum but assuming a product of two jumps must still jump. (D) comes from pairing the wrong sides (left of $f$ with right of $g$ ): that pairing gives $\frac33=\frac{-1}{-1}=1$ for III and different answers for I and II. (C) comes from extending the pattern without checking III.

Where students lose points

Pairing the wrong sides, or concluding "does not exist" for every combination because $f$ and $g$ each jump. Write a two-column table, left and right, compute each combination in its own column, and compare.

original problems, each with a full worked solution: 559; original problems, each with a full worked solution
at test level or beyond: 387; at test level or beyond
short lessons, one for every idea: 43; short lessons, one for every idea
units, all open now: 8; units, all open now

Free lesson

Lesson 1.1: What a Limit Means and How to Read One

Why calculus needs limits, how to read one-sided notation, and how to find limits of single functions, sums, products and compositions straight from a graph or a piecewise formula. Covers College Board AP Calculus AB topics 1.1 to 1.3.

This lesson shows why calculus starts with limits, how to read limit notation including one-sided limits, and how to find limits straight from a graph or a piecewise formula.

Problem 2

Core
True or false
No calculator
2 pts

True or False: A function $g$ can satisfy both $\displaystyle\lim_{x\to 3}g(x)=-1$ and $g(3)=8$ . Justify your answer.

Answer

True. A limit at $3$ is decided only by the values of $g(x)$ for $x$ near $3$ with $x\neq3$ , so the single value $g(3)$ has no say. For example, let $g(x)=2-x$ for $x\neq3$ and $g(3)=8$ . Then $g(x)$ heads to $2-3=-1$ from both sides, so $\lim_{x\to3}g(x)=-1$ , while $g(3)=8$ .

Full solution

Start from what the notation says. $\lim_{x\to3}g(x)=-1$ means you can make $g(x)$ as close to $-1$ as you like by taking $x$ close enough to $3$ , but not equal to $3$ . The phrase in bold is the key to the whole question.
A statement that says something can happen is proved true by one example. So build one.
Begin with a simple function that heads to $-1$ at $x=3$ : the line $y=2-x$ works, since $2-3=-1$ .
Now overrule the single point: define $g(x)=2-x$ for every $x\neq3$ , and set $g(3)=8$ . On a graph, that is the line with an open circle at $(3,-1)$ and a filled dot floating up at $(3,8)$ .
Check the limit from both sides. At $x=2.99$ , $g(x)=-0.99$ ; at $x=3.01$ , $g(x)=-1.01$ . Both sides close in on $-1$ , the left and right limits agree, and so $\lim_{x\to3}g(x)=-1$ . The dot at height $8$ was never used, because $x=3$ itself is excluded.
Check the value: $g(3)=8$ by definition. Both conditions hold at once, so the statement is true.
The big idea: the limit is where the graph is heading; the function value is where the dot is. They match only when the function is continuous at that point, which you will study later in this unit.

Where students lose points

Answering False because "the limit is what you get when you plug in." Plugging in only works for continuous functions. For a limit, cover the point $x=3$ with your finger and read only the nearby values on each side.

Problem 3

Core
Multiple choice
No calculator
4 pts

Which statement means exactly the same thing as $\displaystyle\lim_{x\to-3^{+}}r(x)=6$ ?

Open the rest of this lesson, free

The full lesson and all 13 of its problems, with every solution.

How it works

Class prep that works alongside class

A full AP Calculus AB course for grades 9 to 12, built for the student who is taking it now. It works alongside a school course and also stands on its own. Students who already get the right answers learn to justify each step and to explain an answer in words. Students who feel behind rebuild a topic with the lesson before they try the problems. This course covers the AP Calculus AB topics, so students taking BC should use the AP Calculus BC course.

Every problem is original: true or false, multiple choice, multi-part and free response, each marked calculator or no calculator.
Each one is marked core, test level or beyond.
Solutions show every step, including how to justify an answer when a problem asks for one.
Each solution ends with the mistake that most often costs points.
Worked examples in every lesson, and a graph or diagram for the problems where a picture makes the idea clear.

What is inside

1Limits and Continuity6 lessons · 78 problems
2Differentiation: Definition and Fundamental Properties6 lessons · 78 problems
3Differentiation: Composite, Implicit, and Inverse Functions6 lessons · 78 problems
4Contextual Applications of Differentiation6 lessons · 78 problems
5Analytical Applications of Differentiation6 lessons · 78 problems
6Integration and Accumulation of Change4 lessons · 52 problems
7Differential Equations4 lessons · 52 problems
8Applications of Integration5 lessons · 65 problems

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Who built it

Neil Kothari

TJHSST alumnus · Co-founder, EduAvenues · 10+ years in TJ preparation

Neil is a TJ alumnus and has spent more than a decade working with TJ students and their families. He built this course so students can practice computing and explaining calculus together. Every problem is written for the course and every solution is checked before it goes in.

Tuition

$395for the full school year

One tuition covers every unit of the course. All 8 units are open now. The What a Limit Means and How to Read One lesson stays free for everyone.

All 8 units: 43 lessons and 559 original problems
A full worked solution for every problem
Access through the end of the school year

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